Question

我有三个#include<iostream> #include<map> #include <cmath> #include <cstdio> #include <vector> #include <iostream> #include <algorithm> using namespace std; int main() { int n = 2; vector <vector<int>*> vec1; for (int i = 0; i < n; i++) { vector<int> *vec = new vector<int>; vec->clear(); cout << "Enter number of var : "; int var; cin >> var; for (int i = 0; i < var; i++) { cout << "Enter number : "; int num ; cin >> num; vec->push_back(num); } vec1.push_back(vec); cout << "Container ends \n"; } cout << "Enter i : "; int i; cin >> i; cout << "Enter j : "; int j; cin >> j; auto newvec = vec1[i]; cout << newvec->at(j) << endl;; system("pause"); return 0; }，第一个是基本DF，第二个是行为df，第三个是规则df

df

我想计算每个人的奖励积分，并在奖励名称中添加基本df列，并检查行为df中的条件。如果a和b为true，则将分配100点；如果OR为b，则为b。正确，因此将分配200分，否则将在规则表中a或b条件的地方分配0分

Base df:
+---+----+------+
| ID|Name|Salary|
+---+----+------+
|  1|   A|   100|
|  2|   B|   200|
|  3|   C|   300|
|  4|   D|  1000|
|  5|   E|   500|
+---+----+------+

Behavior DF:
+----+---------+------+
|S.NO|Operation|Points|
+----+---------+------+
|   1|  a AND b|   100|
|   2|   a OR b|   200|
|   3|otherwise|     0|
+----+---------+------+

Rule DF:
+----+-----+------+------------+-----+
|RULE|Table|   col|   operation|value|
+----+-----+------+------------+-----+
|   a| Base|Salary|       equal| 1000|
|   b| Base|Salary|Greater Than|  500|
+----+-----+------+------------+-----+

Answer 1

您可以采用这种方法-

我必须在Rule和Behavior数据帧中进行一些更改。将操作存储为逻辑（“ ==”）而不是字符串（“ equal”）。

Base = spark.createDataFrame([(1,'A',100),(2,'B',200),(3,'C',300),(4,'D',1000),(5,'E',500)],['ID','Name','Salary'])

Behavior = spark.createDataFrame([(1,'df.rule_a & df.rule_b',100),(2,'df.rule_a | df.rule_b',200),(3,'otherwise',0)],['SNo','Operation','Points'])

Rule = spark.createDataFrame([(1,'Base','Salary','==',1000),(2,'Base','Salary','>',500)],['RULE','Table','col','operation','value'])

Base.show()

#+---+----+------+
#| ID|Name|Salary|
#+---+----+------+
#|  1|   A|   100|
#|  2|   B|   200|
#|  3|   C|   300|
#|  4|   D|  1000|
#|  5|   E|   500|
#+---+----+------+

Behavior.show()

#+---+---------------------+------+
#|SNo|           Operation |Points|
#+---+---------------------+------+
#|  1|df.rule_a & df.rule_b|   100|
#|  2|df.rule_a | df.rule_b|   200|
#|  3|           otherwise |     0|
#+---+---------------------+------+

Rule.show()

#+----+-----+------+---------+-----+
#|RULE|Table|   col|operation|value|
#+----+-----+------+---------+-----+
#|   1| Base|Salary|       ==| 1000|
#|   2| Base|Salary|        >|  500|
#+----+-----+------+---------+-----+

为存储在Rules数据框中的规则准备逻辑

要动态地准备规则，可以在for loop数据帧上运行Rule，并传递迭代编号以过滤转换和Rule变量。

from pyspark.sql.functions import  col,concat,lit

rule_a = Rule.filter("RULE == 1").select(concat(col("Table"), lit("['"),  col("col"), lit("']"), lit(" "),  col("Operation"), col("Value"))).collect()[0][0]

rule_b = Rule.filter("RULE == 2").select(concat(col("Table"), lit("['"),  col("col"), lit("']"), lit(" "),  col("Operation"), col("Value"))).collect()[0][0]

将规则执行的布尔结果添加到数据框

df = Base.withColumn("rule_a", eval(rule_a)).withColumn("rule_b", eval(rule_b))

df.show()

#+---+----+------+------+------+
#| ID|Name|Salary|rule_a|rule_b|
#+---+----+------+------+------+
#|  1|   A|   100| false| false|
#|  2|   B|   200| false| false|
#|  3|   C|   300| false| false|
#|  4|   D|  1000|  true|  true|
#|  5|   E|   500| false| false|
#+---+----+------+------+------+

将行为和从Behavior数据帧到变量的对应点存储

要动态准备变量，可以在Behavior数据帧上运行for loop，并将迭代次数作为变量传递，以过滤转换和列名。

behavior1 = Behavior.filter("SNo==1").select( col("Operation")).collect()[0][0]
behavior1_points = Behavior.filter("SNo==1").select( col("Points")).collect()[0][0]

behavior2 = Behavior.filter("SNo==2").select( col("Operation")).collect()[0][0]
behavior2_points = Behavior.filter("SNo==2").select( col("Points")).collect()[0][0]

behavior3 = Behavior.filter("SNo==3").select( col("Operation")).collect()[0][0]
behavior3_points = Behavior.filter("SNo==3").select( col("Points")).collect()[0][0]

最终解决方案

from pyspark.sql.functions import lit,when,col,greatest 

df\
  .withColumn("b1", eval(behavior1))\
  .withColumn("b2", eval(behavior2))\
           .select('*'
                   ,greatest(when(col('b1') == 'true',lit(behavior1_points)).otherwise(0)
                             ,when(col('b2') == 'true',lit(behavior2_points)).otherwise(0)
                             ,lit(behavior3_points)).alias('point')).drop('rule_a','rule_b','b1','b2').show()

#+---+----+------+-----+
#| ID|Name|Salary|point|
#+---+----+------+-----+
#|  1|   A|   100|    0|
#|  2|   B|   200|    0|
#|  3|   C|   300|    0|
#|  4|   D|  1000|  200|
#|  5|   E|   500|    0|
#+---+----+------+-----+

如何从另一个数据框检查布尔条件

1 个答案: